Graphing quadratic functions in factored form worksheet with four problems, including equations, vertex analysis, and graphing exercises.

Problem: Graphing Quadratic Functions in Factored Form Worksheet #2

We are tasked with analyzing and graphing quadratic functions given in factored form. Let's solve each problem step by step.

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Problem 1: \( y = 2(x + 3)(x - 1) \)

#### Step 1: Determine whether the parabola opens up or down
The coefficient of \( x^2 \) (the leading coefficient) is \( 2 \). Since \( 2 > 0 \), the parabola opens up.

#### Step 2: Identify the vertex (Max or Min)
Since the parabola opens up, the vertex is a minimum.

#### Step 3: Find the \( x \)-intercepts
The \( x \)-intercepts occur where \( y = 0 \):
\[
2(x + 3)(x - 1) = 0
\]
Set each factor to zero:
\[
x + 3 = 0 \quad \text{or} \quad x - 1 = 0
\]
\[
x = -3 \quad \text{or} \quad x = 1
\]
So, the \( x \)-intercepts are \( (-3, 0) \) and \( (1, 0) \).

#### Step 4: Find the axis of symmetry
The axis of symmetry is the vertical line passing through the midpoint of the \( x \)-intercepts:
\[
x = \frac{-3 + 1}{2} = \frac{-2}{2} = -1
\]
So, the axis of symmetry is \( x = -1 \).

#### Step 5: Find the vertex
The vertex lies on the axis of symmetry \( x = -1 \). Substitute \( x = -1 \) into the equation to find \( y \):
\[
y = 2((-1) + 3)((-1) - 1)
\]
\[
y = 2(2)(-2) = 2 \cdot 2 \cdot (-2) = -8
\]
So, the vertex is \( (-1, -8) \).

#### Step 6: Determine the domain and range
- Domain: All real numbers, \( (-\infty, \infty) \).
- Range: Since the vertex is a minimum and the parabola opens up, the range is \( [-8, \infty) \).

#### Final Answers for Problem 1:
- Opens Up or Down? Up
- Is the vertex a Max or Min? Min
- \( x \)-intercepts: \( (-3, 0) \) and \( (1, 0) \)
- Axis of symmetry: \( x = -1 \)
- Vertex: \( (-1, -8) \)
- Domain: \( (-\infty, \infty) \)
- Range: \( [-8, \infty) \)

---

Problem 2: \( y = -\frac{1}{2}(x - 4)(x - 8) \)

#### Step 1: Determine whether the parabola opens up or down
The coefficient of \( x^2 \) is \( -\frac{1}{2} \). Since \( -\frac{1}{2} < 0 \), the parabola opens down.

#### Step 2: Identify the vertex (Max or Min)
Since the parabola opens down, the vertex is a maximum.

#### Step 3: Find the \( x \)-intercepts
The \( x \)-intercepts occur where \( y = 0 \):
\[
-\frac{1}{2}(x - 4)(x - 8) = 0
\]
Set each factor to zero:
\[
x - 4 = 0 \quad \text{or} \quad x - 8 = 0
\]
\[
x = 4 \quad \text{or} \quad x = 8
\]
So, the \( x \)-intercepts are \( (4, 0) \) and \( (8, 0) \).

#### Step 4: Find the axis of symmetry
The axis of symmetry is the vertical line passing through the midpoint of the \( x \)-intercepts:
\[
x = \frac{4 + 8}{2} = \frac{12}{2} = 6
\]
So, the axis of symmetry is \( x = 6 \).

#### Step 5: Find the vertex
The vertex lies on the axis of symmetry \( x = 6 \). Substitute \( x = 6 \) into the equation to find \( y \):
\[
y = -\frac{1}{2}(6 - 4)(6 - 8)
\]
\[
y = -\frac{1}{2}(2)(-2) = -\frac{1}{2} \cdot 2 \cdot (-2) = 2
\]
So, the vertex is \( (6, 2) \).

#### Step 6: Determine the domain and range
- Domain: All real numbers, \( (-\infty, \infty) \).
- Range: Since the vertex is a maximum and the parabola opens down, the range is \( (-\infty, 2] \).

#### Final Answers for Problem 2:
- Opens Up or Down? Down
- Is the vertex a Max or Min? Max
- \( x \)-intercepts: \( (4, 0) \) and \( (8, 0) \)
- Axis of symmetry: \( x = 6 \)
- Vertex: \( (6, 2) \)
- Domain: \( (-\infty, \infty) \)
- Range: \( (-\infty, 2] \)

---

Problem 3: \( y = 3x^2 + 12x \)

#### Step 1: Factor the quadratic function
Factor out the greatest common factor (GCF):
\[
y = 3x(x + 4)
\]

#### Step 2: Determine whether the parabola opens up or down
The coefficient of \( x^2 \) is \( 3 \). Since \( 3 > 0 \), the parabola opens up.

#### Step 3: Identify the vertex (Max or Min)
Since the parabola opens up, the vertex is a minimum.

#### Step 4: Find the \( x \)-intercepts
The \( x \)-intercepts occur where \( y = 0 \):
\[
3x(x + 4) = 0
\]
Set each factor to zero:
\[
3x = 0 \quad \text{or} \quad x + 4 = 0
\]
\[
x = 0 \quad \text{or} \quad x = -4
\]
So, the \( x \)-intercepts are \( (0, 0) \) and \( (-4, 0) \).

#### Step 5: Find the axis of symmetry
The axis of symmetry is the vertical line passing through the midpoint of the \( x \)-intercepts:
\[
x = \frac{0 + (-4)}{2} = \frac{-4}{2} = -2
\]
So, the axis of symmetry is \( x = -2 \).

#### Step 6: Find the vertex
The vertex lies on the axis of symmetry \( x = -2 \). Substitute \( x = -2 \) into the equation to find \( y \):
\[
y = 3(-2)^2 + 12(-2)
\]
\[
y = 3(4) + 12(-2) = 12 - 24 = -12
\]
So, the vertex is \( (-2, -12) \).

#### Step 7: Determine the domain and range
- Domain: All real numbers, \( (-\infty, \infty) \).
- Range: Since the vertex is a minimum and the parabola opens up, the range is \( [-12, \infty) \).

#### Final Answers for Problem 3:
- Factor: \( y = 3x(x + 4) \)
- Opens Up or Down? Up
- Is the vertex a Max or Min? Min
- \( x \)-intercepts: \( (0, 0) \) and \( (-4, 0) \)
- Axis of symmetry: \( x = -2 \)
- Vertex: \( (-2, -12) \)
- Domain: \( (-\infty, \infty) \)
- Range: \( [-12, \infty) \)

---

Problem 4: \( y = x^2 + 4x - 12 \)

#### Step 1: Factor the quadratic function
Factor the quadratic expression:
\[
y = x^2 + 4x - 12
\]
Find two numbers that multiply to \( -12 \) and add to \( 4 \). These numbers are \( 6 \) and \( -2 \):
\[
y = (x + 6)(x - 2)
\]

#### Step 2: Determine whether the parabola opens up or down
The coefficient of \( x^2 \) is \( 1 \). Since \( 1 > 0 \), the parabola opens up.

#### Step 3: Identify the vertex (Max or Min)
Since the parabola opens up, the vertex is a minimum.

#### Step 4: Find the \( x \)-intercepts
The \( x \)-intercepts occur where \( y = 0 \):
\[
(x + 6)(x - 2) = 0
\]
Set each factor to zero:
\[
x + 6 = 0 \quad \text{or} \quad x - 2 = 0
\]
\[
x = -6 \quad \text{or} \quad x = 2
\]
So, the \( x \)-intercepts are \( (-6, 0) \) and \( (2, 0) \).

#### Step 5: Find the axis of symmetry
The axis of symmetry is the vertical line passing through the midpoint of the \( x \)-intercepts:
\[
x = \frac{-6 + 2}{2} = \frac{-4}{2} = -2
\]
So, the axis of symmetry is \( x = -2 \).

#### Step 6: Find the vertex
The vertex lies on the axis of symmetry \( x = -2 \). Substitute \( x = -2 \) into the equation to find \( y \):
\[
y = (-2)^2 + 4(-2) - 12
\]
\[
y = 4 - 8 - 12 = -16
\]
So, the vertex is \( (-2, -16) \).

#### Step 7: Determine the domain and range
- Domain: All real numbers, \( (-\infty, \infty) \).
- Range: Since the vertex is a minimum and the parabola opens up, the range is \( [-16, \infty) \).

#### Final Answers for Problem 4:
- Factor: \( y = (x + 6)(x - 2) \)
- Opens Up or Down? Up
- Is the vertex a Max or Min? Min
- \( x \)-intercepts: \( (-6, 0) \) and \( (2, 0) \)
- Axis of symmetry: \( x = -2 \)
- Vertex: \( (-2, -16) \)
- Domain: \( (-\infty, \infty) \)
- Range: \( [-16, \infty) \)

---

Final Answer:

\[
\boxed{
\begin{array}{ll}
\text{Problem 1:} & \text{Opens Up, Min, } x\text{-intercepts: } (-3, 0) \text{ and } (1, 0), \text{ Axis of symmetry: } x = -1, \text{ Vertex: } (-1, -8), \text{ Domain: } (-\infty, \infty), \text{ Range: } [-8, \infty) \\
\text{Problem 2:} & \text{Opens Down, Max, } x\text{-intercepts: } (4, 0) \text{ and } (8, 0), \text{ Axis of symmetry: } x = 6, \text{ Vertex: } (6, 2), \text{ Domain: } (-\infty, \infty), \text{ Range: } (-\infty, 2] \\
\text{Problem 3:} & \text{Factor: } y = 3x(x + 4), \text{ Opens Up, Min, } x\text{-intercepts: } (0, 0) \text{ and } (-4, 0), \text{ Axis of symmetry: } x = -2, \text{ Vertex: } (-2, -12), \text{ Domain: } (-\infty, \infty), \text{ Range: } [-12, \infty) \\
\text{Problem 4:} & \text{Factor: } y = (x + 6)(x - 2), \text{ Opens Up, Min, } x\text{-intercepts: } (-6, 0) \text{ and } (2, 0), \text{ Axis of symmetry: } x = -2, \text{ Vertex: } (-2, -16), \text{ Domain: } (-\infty, \infty), \text{ Range: } [-16, \infty)
\end{array}
}
\]